An electron tube was sealed off during manufacture at a pressure of `1.2xx10^(-7)` mm of mercury at `27^(@)C.` Its volume is 100 `cm^(3)`. The number of molecules that remain in the tube is

Options

A. `2xx10^(16)`
B. `3xx10^(15)`
C. `3.86xx10^(11)`
D. `5xx10^(11)`

Explanation

To find the number of molecules in the electron tube, we will use the ideal gas equation in the form of \( n = \frac{PV}{kT} \), where:

- \( P \) = pressure in Pascals
- \( V \) = volume in cubic meters
- \( k \) = Boltzmann constant \( = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

1. **Convert the given pressure from mm of mercury to Pascals:**
\[ P = 1.2 \times 10^{-7} \, \text{mm Hg} \]
To convert mm Hg to Pascals, we use the conversion factor \( 1 \, \text{mm Hg} = 133.322 \, \text{Pa} \):
\[ P = 1.2 \times 10^{-7} \, \text{mm Hg} \times 133.322 \, \text{Pa/mm Hg} = 1.598 \times 10^{-5} \, \text{Pa} \]

2. **Convert the volume from cm³ to m³:**
\[ V = 100 \, \text{cm}^3 = 100 \times 10^{-6} \, \text{m}^3 = 1.0 \times 10^{-4} \, \text{m}^3 \]

3. **Convert the temperature from Celsius to Kelvin:**
\[ T = 27 \, \text{°C} + 273 = 300 \, \text{K} \]

4. **Substitute the values into the ideal gas equation:**
\[ n = \frac{PV}{kT} \]
Substituting the values we calculated:
\[ n = \frac{(1.598 \times 10^{-5} \, \text{Pa})(1.0 \times 10^{-4} \, \text{m}^3)}{(1.38 \times 10^{-23} \, \text{J/K})(300 \, \text{K})} \]

5. **Calculate the numerator:**
\[ \text{Numerator} = 1.598 \times 10^{-5} \times 1.0 \times 10^{-4} = 1.598 \times 10^{-9} \]

6. **Calculate the denominator:**
\[ \text{Denominator} = 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21} \]

7. **Calculate \( n \):**
\[ n = \frac{1.598 \times 10^{-9}}{4.14 \times 10^{-21}} \approx 3.86 \times 10^{11} \]

Thus, the number of molecules that remain in the tube is approximately \( 3.86 \times 10^{11} \).